Lesson

As a conic section, the parabola fits between the ellipse and the hyperbola.

It is the boundary between the open and closed path of a comet as it is pulled toward the Sun. If the comet is too slow, it becomes trapped into an elliptical orbit - too fast and it escapes along a hyperbolic trajectory.

It is the path taken by a cannon ball when fired out of a cannon at some particular angle of elevation. It is the path of a stream of water as it pushes out of a hose fitting. In fact the parabolic trajectory is a universal one - a stone thrown by someone on the surface of the moon follows a parabola.

It is the natural shape formed by the cable on the Brooklyn Bridge and the Golden Gate Bridge in San Francisco.

The reflective property is perhaps the most scientifically utilised feature of the parabola. Radio telescopes and lighthouses work because of it. Torches and car headlights use parabolic mirrored surfaces that reflect light rays in such a way as to prevent the light from scattering too much. Concrete constructions of parabolic surfaces were even used in England in World War 2 to bring the reflection of distant sounds of enemy aircraft to a single point, so they could be heard.

The property can be stated as follows:

Light rays emanating from the focus of a parabola will strike the inside surface (assuming its made of a mirrored compound) and be reflected back along lines parallel to the parabola's axis of symmetry.

The reflection property is illustrated in the following diagram.

The light ray shown in blue highlights the Law of Reflection. The law guarantees that the two angles $SPN$`S``P``N` and $NPQ$`N``P``Q` measured against the normal $PN$`P``N` as the incident angle $i^{\circ}$`i`∘ and the reflection angle $r^{\circ}$`r`∘ are equal irrespective of where the ray strikes.

While not a complete proof, the following investigation points the way to an understanding of why the parabola exhibits this property. We'll use an example to illustrate to concept.

The parabola with equation given by $x^2=36y$`x`2=36`y` opens upward. Its vertex is located at the origin and the focal length can be found by solving $4a=36$4`a`=36. With $a=9$`a`=9 the focus $S$`S` has coordinates $\left(0,9\right)$(0,9) as shown in the diagram below.

We can easily verify that $P$`P`, with coordinates $\left(12,4\right)$(12,4) is a point on the parabola.

By calculus methods we can show that the gradient of the tangent $TPR$`T``P``R` at $P$`P` is given by $m=\frac{2}{3}$`m`=23 and therefore the equation of the tangent becomes $y-4=\frac{2}{3}\left(x-12\right)$`y`−4=23(`x`−12) which, when simplified, becomes $y=\frac{2}{3}x-4$`y`=23`x`−4.

This tangent cuts the $y$`y` axis at $\left(0,-4\right)$(0,−4).

The distance $ST=9+4=13$`S``T`=9+4=13, as is the distance $SP=\sqrt{\left(12-0\right)^2+\left(4-9\right)^2}=13$`S``P`=√(12−0)2+(4−9)2=13. This means that the triangle $SPT$`S``P``T` is isosceles with base angles equal (shown as $\theta$`θ` in the diagram).

Now here's the punchline.

We know from the Law of Reflection that angles $SPN=i^{\circ}$`S``P``N`=`i`∘ and $NPQ=r^{\circ}$`N``P``Q`=`r`∘ are equal. This implies that the two adjacent angles $SPT$`S``P``T` and $QPR$`Q``P``R` are also equal also. Hence, since angle $SPT$`S``P``T` is $\theta$`θ`, then angle $QPR$`Q``P``R` is also $\theta$`θ`.

But this implies that reflected ray $PQ$`P``Q` must be parallel to the axis of the parabola because of the equal corresponding angles $QPR$`Q``P``R` and $STP$`S``T``P`.

This establishes that the reflective property holds for the *single ray* $SP$`S``P` emanating from the focus of the parabola given by $x^2=36y$`x`2=36`y`.

In the above investigation, we looked at a particular parabola and a particular point $P$`P`. A complete proof shows that for any parabola $x^2=4ay$`x`2=4`a``y` with focus $S$`S` and for any variable point $P$`P` on it, the tangent $TP$`T``P` becomes the third side of a variable isosceles triangle $STP$`S``T``P`. It thus follows that all reflected rays emanating from the focus must be parallel to the parabola's axis of symmetry.

The reverse is also true. Parallel rays which enter the parabola will strike its surface and reflect back to the focus. This is the principle behind radio telescopes, where incoming signals from outer space are collected after reflection at the focus of the parabola.

A $70$70 metre antenna operates at the Tidbinbilla tracking station. It is the largest steerable parabolic antenna in the Southern Hemisphere. Weighing more than $3000$3000 tonnes it rotates on a film of oil. The reflector surface is made up of $1,272$1,272 aluminium panels with a total surface area of $4,180$4,180 square metres.

The photograph has been overlaid with coordinate axes together with a depiction of a cross sectional parabolic curve.

Assuming that the focal length of the parabola is $27$27 metres and the diameter of the dish is $70$70 metres, we seek an equation for the parabola relative to the coordinate axes and the depth of the dish at the centre.

The form of the equation is given as $x^2=4ay$`x`2=4`a``y`, opening from above with the vertex at the origin. Since the focal length is $27$27, the equation becomes $x^2=108y$`x`2=108`y`.

Because the diameter is $70$70 metres, the point $B$`B` in the photograph must have coordinates given by $\left(35,y_B\right)$(35,`y``B`) where the value of $y_B$`y``B` can be found.

$x^2$x2 |
$=$= | $108y$108y |

$35^2$352 | $=$= | $108\left(y_B\right)$108(yB) |

$1225$1225 | $=$= | $108\left(y_B\right)$108(yB) |

$\therefore$∴ $y_B$yB |
$\approx$≈ | $11.343$11.343 |

This means the depth of the dish at the centre is $11.343$11.343 metres.

A satellite dish is parabolic in shape, with a diameter of $8$8 metres. Incoming signals are reflected to one collection point, the focus of the parabola, marked as point $F$`F` on the diagram (not to scale). The focus is positioned such that the focal length is $4$4 metres.

Using $\left(0,0\right)$(0,0) as the vertex of the parabola, solve for $d$

`d`, the depth of the dish in metres.

A parabolic antenna has a cross-section of width $16$16 m and depth of $2$2 m. All incoming signals reflect off the surface of the antenna and pass through the focus at $F$`F`. Note: Image is not to scale

Placing the vertex of the antenna at $\left(0,0\right)$(0,0), we can model the surface of the antenna by the parabola with equation $y^2=4ax$

`y`2=4`a``x`.State the equation of parabola.

For best reception, at what coordinate should the receiver be placed?

When a fire hose is turned on, the height $h$`h` metres of the stream of water is described by $h=-0.25d\left(d-32\right)$`h`=−0.25`d`(`d`−32), where $d$`d` metres is the horizontal distance from the hose.

Loading Graph...

Let the hose be positioned at $\left(0,0\right)$(0,0). Use the graph to determine how far from the hose the water reaches its maximum height above ground.

$\editable{}$ metres from the hose

Using the graph, determine the maximum height reached by the hose.

There are two points on the path where the water is at a height of $48$48 metres above the ground. What is the horizontal distance between these two points?

Determine the height of the water above the ground when it is a horizontal distance of $6$6 metres from where the fire is.

For what values of $d$

`d`is the model valid?$d\le32$

`d`≤32 and $d\ge0$`d`≥0A$d<0$

`d`<0 and $d<32$`d`<32B$d\ge0$

`d`≥0 and $d>32$`d`>32C$d>32$

`d`>32 and $d<0$`d`<0D$d\le32$

`d`≤32 and $d\ge0$`d`≥0A$d<0$

`d`<0 and $d<32$`d`<32B$d\ge0$

`d`≥0 and $d>32$`d`>32C$d>32$

`d`>32 and $d<0$`d`<0D

Apply the geometry of conic sections

Apply the geometry of conic sections in solving problems